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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #22 : Differentials

Find the differential of the following function:

$f(x,y,z)=\cos^2(xy)+\sin(z)$

Possible Answers:

$-2y\cos(xy)\sin(xy)dx-2x\cos(xy)\sin(xy)dy-\cos(z)dz$

$-y\cos(xy)\sin(xy)dx-x\cos(xy)\sin(xy)dy+\cos(z)dz$

$2y\cos(xy)\sin(xy)dx+2x\cos(xy)\sin(xy)dy+\cos(z)dz$

$-2y\cos(xy)\sin(xy)dx-2x\cos(xy)\sin(xy)dy+\cos(z)dz$

$-2x\cos(xy)\sin(xy)dx-2y\cos(xy)\sin(xy)dy+\cos(z)dz$

Correct answer:

$-2y\cos(xy)\sin(xy)dx-2x\cos(xy)\sin(xy)dy+\cos(z)dz$

Explanation:

The differential of a function f(x,y,z) is given by

f_xdx+f_ydy+f_zdz

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=-2y\cos(xy)\sin(xy)$

$f_y=-2x\cos(xy)\sin(xy)$

$f_z=\cos(z)$

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Example Question #3803 : Calculus 3

Find the differential of the following function:

$f(x,y,z)=7xy+6x^2y^3e^{z^2}$

Possible Answers:

$(7x+12xy^2e^{z^2})dx+(7y+18x^2y^2e^{z^2})dy+(12x^2y^3ze^{z^2})dz$

$(7y+12xy^2e^{z^2})dx+(7x+12x^2y^2e^{z^2})dy+(12x^2y^3ze^{z^2})dz$

$(7y+12xy^2e^{z^2})dx+(7x+18x^2y^2e^{z^2})dy+(6x^2y^3ze^{z^2})dz$

$(7y+12xy^2e^{z^2})dx+(7x+18x^2y^2e^{z^2})dy+(12x^2y^3ze^{z^2})dz$

$(7y+12xy^2e^{z^2})dx+(7x+18x^2y^2e^{z^2})dy+(12x^2y^3z^2e^{z^2})dz$

Correct answer:

$(7y+12xy^2e^{z^2})dx+(7x+18x^2y^2e^{z^2})dy+(12x^2y^3ze^{z^2})dz$

Explanation:

The differential of a function f(x,y,z) is given by

f_xdx+f_ydy+f_zdz

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=7y+12xy^2e^{z^2}$

$f_y=7x+18x^2y^2e^{z^2}$

$f_z=12x^2y^3ze^{z^2}$

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Example Question #23 : Differentials

Find the total differential of the following function:

$f(x,y,z)=xz^4+y\sqrt{z}$

Possible Answers:

$z^4dx+\sqrt{z}dy+(4z^3+\frac{y}{2\sqrt{z}})dz$

$z^4dx+\sqrt{z}dy+(4xz^3+\frac{y}{2\sqrt{z}})dz$

$z^4+\sqrt{z}+4xz^3+\frac{y}{2\sqrt{z}}$

$z^4dx+\sqrt{z}dy+(4xz^3+\frac{y\sqrt{z}}{2})dz$

Correct answer:

$z^4dx+\sqrt{z}dy+(4xz^3+\frac{y}{2\sqrt{z}})dz$

Explanation:

The total differential of a function f(x,y,z) is given by

f_xdx+f_ydy+f_zdz

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

f_x=z^4 , $f_y=\sqrt{z}$ , $f_z=4xz^3+\frac{y}{2\sqrt{z}}$

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Example Question #24 : Differentials

Find the total differential of the function:

$f(x, y, z)=8z\ln(xy)$

Possible Answers:

$\frac{8z}{x}dx+\frac{8z}{y}dy+8\ln(xy)dz$

$\frac{8yz}{x}dx+\frac{8xz}{y}dy+8\ln(xy)dz$

$\frac{8z}{x}+\frac{8z}{y}+8\ln(xy)$

$\frac{8z}{y}dx+\frac{8z}{x}dy+8\ln(xy)dz$

Correct answer:

$\frac{8z}{x}dx+\frac{8z}{y}dy+8\ln(xy)dz$

Explanation:

The total differential of a function f(x,y,z) is given by

f_xdx+f_ydy+f_zdz

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=\frac{8z}{x}$ , $f_y=\frac{8z}{y}$ , $f_z=8\ln(xy)$

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Example Question #26 : Differentials

If $z=f(x,y)=x^6+3xy-\sin(xy)$ , calculate the total differential .

Possible Answers:

$dz=(6x^5+3y-y\cos(xy))dx+(3x-x\cos(xy))dy$

$dz=(6x^5-3y+y\cos(xy))dx+(3+x\cos(xy))dy$

$dz=(6x^5+3y-\cos(xy))dx+(3x-\cos(xy))dy$

$dz=(6x^5+3y+y\cos(xy))dx+(3x+x\cos(xy))dy$

$dz=(\frac{x^7}{7}+3y-y\cos(xy))dx+(\frac{3x^2}{2}-x\cos(xy))dy$

Correct answer:

$dz=(6x^5+3y-y\cos(xy))dx+(3x-x\cos(xy))dy$

Explanation:

The total differential of a function z=f(x,y) is defined as the sum of the partial derivatives of with respect to each of its variables; that is,

$dz=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$

In this case, $z=f(x,y)=x^6+3xy-\sin(xy)$ , and so we use the sum rule, the rule for derivatives of a variable raised to a power, the rule for the derivative of $\sin(x)$ , and the chain rule to calculate the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ , as shown:

$\frac{\partial f}{\partial x}=6x^5+3y-y\cos(xy)$ ,

$\frac{\partial f}{\partial y}=3x-x\cos(xy)$ .

In both cases, we treated the variable not being differentiated as a constant, and applied the chain rule to $\sin(xy)$ to calculate its partial derivatives. Now that $\frac{\partial f}{\partial x}dx$ and $\frac{\partial f}{\partial y}dy$ have been calculated, all that remains is to substitute them into the definition of the total derivative:

$dz=(6x^5+3y-y\cos(xy))dx+(3x-x\cos(xy))dy$

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Example Question #27 : Differentials

If z=f(x,y)=xe^x^y , calculate the total differential .

Possible Answers:

e^x^y(1+xy)+2xe^x^y

e^x^y(1+xy)+x^2e^x^y

e^x^y(1+y)+x^2e^x^y

e^x^y(1+xy)+xe^x^y

e^x^y(1+x)+x^2e^x^y

Correct answer:

e^x^y(1+xy)+x^2e^x^y

Explanation:

The total differential of a function z=f(x,y) is defined as the sum of the partial derivatives of with respect to each of its variables; that is,

$dz=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$

In this case, z=f(x,y)=xe^x^y , and so we use the product rule and the rule for differentiating e^x to calculate the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ , as shown:

$\frac{\partial f}{\partial x}=(1)(e^x^y)+(x)(ye^x^y)=e^x^y+xye^x^y=e^x^y(1+xy)$ ,

$\frac{\partial f}{\partial y}=(x)xe^x^y=x^2e^x^y$

In both cases, we treated the variable not being differentiated as a constant, and applied the product rule to calculate its partial derivatives. Now that $\frac{\partial f}{\partial x}dx$ and $\frac{\partial f}{\partial y}dy$ have been calculated, all that remains is to substitute them into the definition of the total derivative:

dz=e^x^y(1+xy)dx+(x^2e^x^y)dy

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Example Question #1447 : Partial Derivatives

Compute the differentials for the following function.

$z=e^{x^2+4y^2}\sin(y)$

Possible Answers:

$dz=8y\cos(y)e^{x^2+4y^2} dy$

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

$dz=2xe^{x^2+4y^2}\sin(y)+8y\cos(y)e^{x^2+4y^2}$

$dz=2xe^{x^2+4y^2}\sin(y)$

$dz=2xe^{x^2+4y^2}\sin(y)dx$

Correct answer:

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

Explanation:

What we need to do is take derivatives, and remember the general equation.

$z=e^{x^2+4y^2}\sin(y)$

w=g(x,y,z)

dw=g_xdx+g_ydy+g_zdz

When taking the derivative with respect to y recall that the product rule needs to be used.

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

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Example Question #1441 : Partial Derivatives

Compute the differentials for the following function.

$z=e^{x^2+4y^2}\sin(y)$

Possible Answers:

$dz=2xe^{x^2+4y^2}\sin(y)$

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

$dz=2xe^{x^2+4y^2}\sin(y)dx$

$dz=8y\cos(y)e^{x^2+4y^2} dy$

$dz=2xe^{x^2+4y^2}\sin(y)+8y\cos(y)e^{x^2+4y^2}$

Correct answer:

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

Explanation:

What we need to do is take derivatives, and remember the general equation.

$z=e^{x^2+4y^2}\sin(y)$

w=g(x,y,z)

dw=g_xdx+g_ydy+g_zdz

When taking the derivative with respect to y recall that the product rule needs to be used.

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #22 : Differentials

Example Question #3803 : Calculus 3

Example Question #23 : Differentials

Example Question #24 : Differentials

Example Question #26 : Differentials

Example Question #27 : Differentials

Example Question #1447 : Partial Derivatives

Example Question #1441 : Partial Derivatives

All Calculus 3 Resources

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